Research Methods and Statistics for Psychology: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit

2Practice problems

1.Research methods in psychologyMedium

Experimenter bias is reduced most directly by:

  1. AUsing only volunteers
  2. BUsing a one-tailed hypothesis
  3. CUsing a larger sample
  4. DUsing a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychologyHard

A test of "intelligence" measures mainly how fast people can type. Which type of validity is most clearly lacking?

  1. AInternal consistency
  2. BInter-rater reliability
  3. CConstruct validity
  4. DTest-retest reliability
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 39, 11, 41, 54, 57.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 70?

μ = 100σ = 20x = 70
P =
5.ProbabilityMedium
xp(x)00.3510.3520.130.2μ = 1.15

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3p(x): 0.35, 0.35, 0.1, 0.2
Var(X) =
6.ProbabilityHard
xp(x)00.2510.120.3530.240.1μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.1, 0.35, 0.2, 0.1Y = 0.5X + 1
σY =
7.Normal distributionMedium
23.343.1P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23.3 and 43.1 min? Give P as a decimal.

μ = 32σ = 6x1 = 23.3x2 = 43.1
P =
8.Normal distributionHard
235240245250255260265gtop 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
9.Confidence intervalsMedium
n = 6, s = 7.5LU = ?x̄ = 115.5E99%

A sample of 6 items has mean 115.5 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 6x̄ = 115.5s = 7.599%t(5) = 4.032
upper limit =
10.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 38. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?

σ = 38E = 0.590%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

20 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.14 and the standard deviation of the differences is 3. Find the paired test statistic t for H₀: μd = 0.

n = 20d̄ = 1.14sd = 3
t =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 390%
n =
13.Linear regression & correlationMedium
xy127221322418514610xy2461020

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy121220319412510xy24101520x = 4.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 7) is made. Find its residual e = y − ŷ.

n = 5(4.5, 7)
e =
15.Chi-square testsMedium
busbikecarTotalstudents175647120staff315237120Total4810884240

The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.

N = 2402 × 3
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: D · Using a researcher who does not know the hypothesis or conditions to collect data

Why this is the answer

Experimenters who know the hypothesis may unintentionally influence participants or recordings. Single- or double-blind procedures and standardised instructions reduce this.

Answer

D · Using a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychology

Answer: C · Construct validity

Why this is the answer

Construct validity is whether a measure truly reflects the underlying construct. A test can be reliable while measuring the wrong thing.

Answer

C · Construct validity
3.Descriptive statistics

Answer: s = 18.22

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 202/5 = 40.4

Squared deviations

Σ(x − x̄)² = (−1.4)² + (−29.4)² + 0.6² + 13.6² + 16.6² = 1327

Standard deviation

s = √(1327/4) = 18.22

Answer

s = 18.22
4.Descriptive statistics

Answer: P = 0.9332

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
5.Probability

Answer: Var(X) = 1.227

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.35 + 1·0.35 + 2·0.1 + 3·0.2 = 1.15

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.1 + 9·0.2 = 2.55

Variance

Var(X) = 2.55 − 1.15² = 1.227

Answer

Var(X) = 1.227
6.Probability

Answer: σY = 0.6442

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Spread of X

μ = 1.8, Var(X) = 1.66, σX = 1.288

Linear change

σY = |0.5|·1.288 = 0.6442

Answer

σY = 0.6442

The mean does move: E(Y) = 0.5·1.8 + 1 = 1.9.

7.Normal distribution

Answer: P = 0.8943

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (23.3 − 32)/6 = −1.45
z₂ = (43.1 − 32)/6 = 1.85

Area between

P = Φ(1.85) − Φ(−1.45) = 0.9678 − 0.07353 = 0.8943

Answer

P = 0.8943
8.Normal distribution

Answer: x = 251.3 g

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g
9.Confidence intervals

Answer: upper limit = 127.8

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Critical value

df = n − 1 = 5, t = 4.032

Margin of error

E = 4.032·7.5/√6 = 12.35

Interval

115.5 ± 12.35 → [103.2, 127.8]

Answer

upper limit = 127.8
10.Confidence intervals

Answer: n = 15631

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (1.645·38/0.5)² = 15630

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 15631

Answer

n = 15631
11.Hypothesis tests & confidence intervals

Answer: t = 1.699

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error of the mean difference

SE = 3/√20 = 0.6708

Test statistic

t = 1.14/0.6708 = 1.699

Answer

t = 1.699

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 20

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Solve for n

n = (1.645·8/3)² = 19.24

Round up

n = 20

Answer

n = 20
13.Linear regression & correlation

Answer: r = −0.971

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 21/6 = 3.5
ȳ = 112/6 = 18.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −55
Syy = Σ(y − ȳ)² = 183.3

Correlation coefficient

r = −55/√(17.5·183.3) = −0.971

Answer

r = −0.971
14.Linear regression & correlation

Answer: e = −4.9

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 15/5 = 3
ȳ = 82/5 = 16.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 16.4 − (−3)·3 = 25.4
ŷ = 25.4 − 3·x

Predicted value

ŷ = 25.4 + (−3)·4.5 = 11.9

Residual

e = 7 − 11.9 = −4.9

Answer

e = −4.9

The point lies below the line.

15.Chi-square tests

Answer: χ² = 5.422

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Expected counts

students: 120·48/240 = 24, 120·108/240 = 54, 120·84/240 = 42
staff: 120·48/240 = 24, 120·108/240 = 54, 120·84/240 = 42

Chi-square statistic

χ² = (17 − 24)²/24 + (56 − 54)²/54 + (47 − 42)²/42 + (31 − 24)²/24 + (52 − 54)²/54 + (37 − 42)²/42
χ² = 5.422

Answer

χ² = 5.422

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 6

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Degrees of freedom

df = 7 − 1 = 6

Answer

df = 6

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